# Parallel and series arrangement

1. Jul 14, 2015

### Lim Y K

Effective resistance in parallel is 1/r1+1/r2+1/r3. However, i don't get why there is lesser resistance when the bulbs are arranged in parallel. I thought arranging a circuit in parallel requires more wire(conductors) , which will increase the resistivity. can someone explain to me please? thank you in advance

2. Jul 14, 2015

### Staff: Mentor

There's more wire involved, but the various paths through the network don't have to travel through all of it so the resistance doesn't have to higher. If you work out the $1/R_1+1/R_2$ formula for yourself you'll see how this works:
- What is the current through R1?
- What is the current through R2?
- What is the total current through the network, and what does that tell you about the effective resistance?

Last edited by a moderator: Jul 14, 2015
3. Jul 14, 2015

### Staff: Mentor

Lengthening a single wire increases total resistance, but making the wire a larger diameter decreases it. By putting the light bulbs in parallel you are kind of doing the latter. As Nugatory said, there are multiple path choices, so any particular charge will only flow through one path and will only encounter resistance equal to one path.

4. Jul 14, 2015

### Lim Y K

I see. Then may I know why the total resistance in a parallel circuit is smaller than any of the individual resistance?

5. Jul 14, 2015

### Staff: Mentor

The voltage applied to each path is the same, and the current through each path is just the voltage divided by the resistance of that path. But when the paths come together the current from each is summed together. So the 'effective' resistance is less than the resistance of any one path. I suggest doing what Nugatory suggested in post #2.

6. Jul 14, 2015

### Staff: Mentor

As I said above, you should be able to derive the formula for yourself - when you do, you'll see why it makes sense. Start with $V=IR$; a trivial rearrangement gives you $R=V/I$ which tells you what the effective resistance is if you know the voltage and the current through the network. $I$ is the sum of the currents through each path, so as you add paths while holding the voltage constant $I$ increases and $R$ decreases.

Don't take my word for it though - work it out for yourself so you know where the formula came from. The basic idea is in my post #2 above.

7. Jul 15, 2015

### Khashishi

To help you conceptualize it, just think of water pipes connected to a water tower or something. Will two pipes in parallel carry more or less water than one pipe? What about a pipe twice as long (but dropping the same vertical distance)?